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SpyIntel [72]
3 years ago
12

Which type of graph would be best for displaying the frequencies of weights of several juvenile monkeys at a wild animal preserv

e according to equal-sized weight categories?
A. a bar graph
B. a circle graph
C. a histogram
D. a line graph
Mathematics
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

C. A histogram

Step-by-step explanation:

Histograms are commonly used to denote the frequency of a range of categories of a set of values. Generally, the y-axis represents frequency and the x-axis represents intervals of data based on the given set of values.

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Show that 3650 is not a perfect square.​
KatRina [158]

Answer:

A number that is a perfect square never ends in 2, 3, 7 or 8. If your number ends in any of those numbers, you can stop here because your number is not a perfect square. Obtain the digital root of the number. The digital root essentially is the sum of all of the digits.

Step-by-step explanation:

hope this helps you super sorry if it does not

4 0
2 years ago
Read 2 more answers
)) Evaluate the expression for b = 7.<br> 11b - 75 =
Wewaii [24]

Answer:

The answer is 2.

Step-by-step explanation:

11(7) - 75

77-75

=2

7 0
3 years ago
Based on what you know now, how are linear and exponential functions alike?
Ainat [17]

Answer:

They're similar in that they both have to maintain a steady rate of rise as they grow. While graphing, you can't adjust the slope or exponent after traveling up a graph.

Step-by-step explanation:

5 0
2 years ago
Help............................
Lyrx [107]

Answer: convex

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8 0
3 years ago
The illuminance of a surface varies inversely with the square of its distance from the light source.If the illuminance of a surf
VashaNatasha [74]

Answer:

Distance between the surface and source of light will be increased by 6 meters.

Step-by-step explanation:

The illuminance of a surface varies inversely with the square of ts distance from the light source.

Let illuminance of the surface = x lumens per square meter

and distance from a light source = y meter.

Now x ∝ \frac{1}{y^{2} }

Or x=\frac{k}{y^{2} } [k = proportionality constant]

Now we will find the value of k.

k = xy²

k = 120×(6)²

k = 4320

We have to calculate the distance of the source if illuminance of the surface is 30 lumens per square meter.

30=\frac{4320}{y^{2}}

y² = \frac{4320}{30}

y² = 144

y = √144 = 12 meters

So the source of the light will be shifted away from the surface = 12 - 6 = 6 meters.

6 0
3 years ago
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